მიმდინარე დრო:0:00მთლიანი ხანგრძლივობა:11:23

ენერგიის 0 ქულა

# Chilling water problem

ვიდეოს აღწერა

Let's do another states of matter phase change problem. And we'll deal with water again. But this one hopefully will stretch our neurons a little bit further. So let's say I have 500 grams of water. Of liquid water. At 60 degrees Celsius. Now my goal is to get it to zero degrees Celsius. And the way I'm going to do it is, I'm going to put ice into this 500 grams of water. And my ice machine at home makes ice that comes out of the machine at minus 10 degrees Celsius ice. And my question is, exactly how much ice do I need? So how much, or how many grams of ice? And I'm going to take the ice out of the freezer and just plop it into my liquid. How much do I need to bring this liquid, this 500 grams of liquid water, down to zero degrees? So the idea, if we just imagine a cup here. Let me draw a cup. This is a cup. I have some 60 degree water in there. I'm going to plunk a big chunk of ice in there. And what's going to happen is that heat from the water is going to go into the ice. So the ice is going to absorb heat from the water. So in order for water to go from 60 degrees to 0 degrees, I have to extract heat out of it. And we're about to figure out just how much heat. And so we have to say, whatever was extracted out of the water, essentially has to be contained by the ice. And the ice can't get above zero degrees. Essentially, that amount of ice has to absorb all that heat to go from minus 10 to zero. And then also, that energy will be used to melt it a bit. But if we don't have enough ice, then the ice is going to go beyond that and then warm up even more. So let's see how we do this. So how much energy do we have to take out of the 500 grams of liquid water? Well, it's the same amount of energy that it would take to put into zero degrees liquid water and get it to 60 degrees. So we're talking about a 50 degree change. So the energy or the heat out of the water is going to be the specific heat of water, And I have to multiply that times the number of grams of water I have to cool down, I have to take the heat out of. And we know that's 500 grams. And then I multiply that times the temperature differential that we care about. And just a side note, I use this specific heat because we're dealing with liquid water. Liquid water going from 60 to zero. So the final thing, I have to multiply it by the change in temperature. The change in temperature is 60 degrees. Times 60 degrees. There's a little button on the side of my pen, when I press it by accident sometimes it does that weird thing. So let's see what this is. So this is 4.178 times 500 times change of 60 degrees. It could be a change of 60 degrees Kelvin or a change of 60 degrees Celsius. It doesn't matter. The actual difference is the same, whether we're doing Kelvin or Celsius. And it's 125,340 joules. So this is the amount of heat that you have to take out of 60 degree water in order to get it down to 0 degrees. Or the amount of heat you have to add to zero degree water to get it to 60 degrees. So essentially, our ice has to absorb this much energy without going above zero degrees. So how much energy can the ice absorb? Well let's set a variable. The question is how much ice. So let's set our variable, maybe we'll call it I. Let's do x. x is always the unknown variable. So we're going to have x grams of ice. OK, and it starts at minus 10 degrees. So when this x grams of ice warms from minus 10 degrees to zero degrees Celsius, how much energy will it be absorbing? So to go from minus 10 degrees Celsius to 0 degrees Celsius the heat that's absorbed by the ice is equal to -- is equal to the specific heat of ice, ice water, times the amount of ice. That's what we're solving for. So times x. Times the change in temperature. So this is a 10 degree change in Celsius degrees, which is also a 10 degree change in Kelvin degrees. We can just do 10 degrees. I could write Kelvin here, just because at least when I wrote the specific heat units, I have a Kelvin in the denominator. It could have been a Celsius, but just to make them cancel out. This is, of course, x grams. So the grams cancel out. So that heat absorbed to go from minus 10 degree ice to zero degree ice is 2.05 times 10 is 20.5. So it's 20.5 times x joules. This is to go from minus 10 degrees to zero degrees. Now, once we're at zero degrees, the ice can even absorb more energy before increasing in temperature as it melts. Remember, when I drew that phase change diagram. The ice gains some energy and then it levels out as it melts. As the the bonds, the hydrogen bonds start sliding past each other, and the crystalline structure breaks down. So this is the amount of energy the ice can also absorb. Let me do it in a different color. Zero degree ice to zero -- I did it again --to zero degree water. Well the heat absorbed now is going to be the heat of fusion of ice. Or the melting heat, either one. That's 333 joules per gram. It's equal to 333.55 joules per gram times the number of grams we have. Once again, that's x grams. They cancel out. So the ice will absorb 333.55 joules as it goes from zero degree ice to zero degree water. Or 333.55x joules. Let me put the x there, that's key. So the total amount of heat that the ice can absorb without going above zero degrees... Because once it's at zero degree water, as you put more heat into it, it's going to start getting warmer again. If the ice gets above zero degrees, there's no way it's going to bring the water down to zero degrees. The water can't get above zero degrees. So how much total heat can our ice absorb? So heat absorbed is equal to the heat it can absorb when it goes from minus 10 to zero degrees ice. And that's 20.5x. Where x is the number of grams of ice we have. Plus the amount of heat we can absorb as we go from zero degree ice to zero degree water. And that's 333.55x. And of course, all of this is joules. So this is the total amount of heat that the ice can absorb without going above zero degrees. Now, how much real energy does it have to absorb? Well it has to absorb all of this 125,340 joules of energy out of the water. Because that's the amount of energy we have to extract from the water to bring it down to zero degrees. So the amount of energy the ice absorbs has to be this 125,340. So that has to be equal to 125,340 joules. We can do a little bit of algebra here. Add these two things. 20.5x plus 333.55x is 354.05x. Is that right? Yeah, 330 plus 30 is 350. Then you have a 3 with a 0.5 there. 354.05x and that is equal to the amount of energy we take out of the water. You divide both sides. So x is equal to 125,340 divided by 354.05. I'll take out the calculator for this. The calculator tells me 125,340, the amount of energy that has to be absorbed by the ice, divided by 354.05 is equal to 354 grams. Roughly, there's a little bit extra. So actually, just to be careful maybe I'll take 355 grams of ice. Because I definitely want my water to be chilled. So our answer is x is equal to 354.02 grams of ice. So this is interesting. I had 500 grams of liquid. And you know, intuitively I said, well if I have to bring that down to zero degrees, I'd have to have a ton of ice. But it turns out, I only need, what was the exact number? So the liquid is 500 grams. About, roughly half a pound, if you want to get a sense for how much 500 grams is. A kilogram is 2.2 pounds. So this ice is 354. So you actually have to have less ice than water. Which is interesting, because it seems like the ice isn't making that big of a temperature change. The ice is only going from minus 10 to zero degrees. While the water is going all the way from 60 degrees to zero degrees. So you're like, how does that work out? And the reason is because so much energy can be absorbed by the ice in the form of potential energy as it melts. So if you talk about all the energy that the ice is asborbing, most of it is due to the heat of fusion right there. You can actually have zero degree ice it can still absorb this huge amount of energy just by melting it. Without even changing its temperature. Anyway, hopefully you found that reasonably useful.